Nasser S. Alzayed

Lecture #7Lecture #7

Nasser S. [email protected]

Chapter 7: Energy BandsKronig‐Penny Model

This model solves for periodic potential in a form of a square‐wellarray:

Schrodinger wave equation for this potential can take the form:

2 2

( ) (11)d U x

2 ( ) (11)2

U xm dx

where U(x) is the potential energy and ε is the energy eigenvalue.W h 2 i 0 < < (U 0) d b < < 0 (U ≠ 0)

King Saud University, Physics Dept. Phys. 570, Nasser S. Alzayed ([email protected])

We have 2 regions: 0 < x < a (U = 0) and ‐b < x < 0 (U ≠ 0)


For the first region we have the wave function:

(12)iKx iKxAe Be This is a combination of plane waves traveling to the right and to the left, with energy:

2 2

= 2Km

(13)

In the 2nd region, wave function takes the form:

(14)Qx QxCe De

2 2

with energy:

= (15)2oQU


( )2o m


Solution of this equation shall be on the Bloch form (7) because itis a periodic potential.

Thus the solution in the region a<x<a+b must be related to the Thus the solution in the region a<x<a+b must be related to thesolution (14) in the region ‐b<x<0 by the Bloch theorem:

.( ) ( ) (7)ik rk kr u r e

The constants A B C D are chosen so that Ψ and Ψ’ are

( )

( ) ( ) (7)

( ) (- 0) (16)k k

ik a b

r u r e

a x a b b x e

The constants A, B, C, D are chosen so that Ψ and Ψ arecontinuous at x=0 and x=a. same as in square potential wells.

At x = 0 we have (for both conditions): (12) + (14):

(17)Derivatives of (12) and (14) at x = 0 provides:A B C D


( - ) ( - ) iK A B Q C D (18)


for the case of x = a; and applying continuity of Ψ and Ψ’ we willget:

( )iK iK Qb Qb ik b( )

( )

( ) (19)( ) ( ) (20)

iKa iKa Qb Qb ik a b

iKa iKa Qb Qb ik a b

Ae Be Ce De eiK Ae Be Q Ce De e

Solving equations from (17) to (20) can be done by putting allcoefficients of A, B, C, D in a determinant. However, such solutionis very difficult We will only write down the final equation:is very difficult. We will only write down the final equation:

2 2

sinh sin cosh cos cos ( ) (21 )2

Q K Qb Ka Qb Ka k a b aQK

2QK

to simplify the solution; we represent the potential by a Delta


Function. Let: b 0 and Uo∞


We do some approximation: Qb 1Q K

sinhcosh 1

( )

Qb QbQb

k b k

2 2

2

cos ( ) cosnotice that: , hence: (21a)

k a b kaQ K

Q

2

sin cos cos2Q Qb Ka Ka kaQKP

2

sin cos cos (21 )P Ka Ka ka bKa

Q bawith P


2with P


The ranges of K for which this equation has solutions are plottedin Fig. 5, for the case P = 3π/2. The corresponding values of theenergy are plotted in Fig 6 Note the energy gaps at the zoneenergy are plotted in Fig. 6. Note the energy gaps at the zoneboundaries.


Chapter 7: Energy BandsWAVE EQUATION OF ELECTRON IN A PERIODIC POTENTIAL

We considered in Fig. 3 the approximate form we expect for thesolution of the Schrödinger equation if the wave vector is at azone boundary as at k = ±π/azone boundary, as at k = ±π/a.

Here, we treat in detail the wave equation for a general potential,at general values of k.

Let U(x) denote the potential energy of an electron in a linearlattice of lattice constant a. Potential energy is invariant underthe lattice translation. Hence: U(x ) = U(x + a).( ) ( )

A function invariant under a crystal lattice translation may beexpanded as a Fourier series in the reciprocal lattice vectors G.

As a rule: There is periodicity : There is Fourier Transform As a rule: There is periodicity : There is Fourier TransformWe write the Fourier series for the potential energy as:

( ) (22)iGxU x U eKing Saud University, Physics Dept. Phys. 570, Nasser S. Alzayed ([email protected])

( ) (22)GGU x U e


Please review Fourier Analysis. For actual crystal potentials; the values of the coefficients UGtend to decrease rapidly with increasing magnitude of G For atend to decrease rapidly with increasing magnitude of G. For acoulomb potential: UG decreases as 1/G2

In equation (22), we did not specify the x‐values. We want to useonly real values. Accordingly: (22) can be rewritten as:

0 0( ) ( ) 2 cos (23)iGx iGx

G GG GU x U e e U Gx

The wave equation of an electron in the crystal is H = whereH is the Hamiltonian and is the energy eigenvalue.

Th f ll ti i th The full equation is then:2 2

( ) ( ) ( ) ( ) ( ) (24)2 2

iGxp pU x x U x e x x


2 2 Gm m


Equation (24) is written in the one‐electron approximation inwhich the orbital (x) describes the motion of one electron in thepotential of the ion cores and in the average potential of thepotential of the ion cores and in the average potential of theother conduction electrons.

The wavefunction (x) may be expressed as a Fourier seriessummed over all values of the wavevector permitted by theboundary conditions, so that:

ik ( ) (25)ikx

kC k e

The set of values of k has the form 2πn/L. To solve the wave equation, substitute (25) in (24) to obtain a setof linear algebraic equations for the Fourier coefficients.

We will take the solution as term by term:


We will take the solution as term by term:


The kinetic energy term is:

22 2 2 221 ikp d d 2

2

1( ) ( ) ( ) ( )2 2 2 2and the potential energy term is :

ikx

k

p d dx i x k C k e im m dx m dx m

( ) ( ) ( ) ( ) ( )

The full Schrodinger Eq

iGx iGx ikxG G

G G kU x x U e x U e C k e ii

becomes:The full Schrodinger Eq. 2

2 ( )

becomes:

( ) ( ) ( ) (26)2

ikx i k G x ikxG

k G k kk C k e U C k e C k e

m

2 2

For 1 K value, this equation becomes:

( ) ( ) ( ) (26)2

ikxG

k C k U C k G C k e


( ) ( ) ( ) ( )2 G

G km

Chapter 7: Energy BandsSolution of the Central Equation

Each Fourier component must have the same coefficient on bothsides of the equation. Thus we have the central equation:

( ) ( ) ( ) 0 (27)

:

k GG

C k U C k G

with

2 2 / 2 (28)k k m

Eq. (27) is the Algebraic form of the well know Schrodingerequation in a periodic potential (24).

It is not easy to solve it, but usually one can use only few terms. No. of solutions of this equation = no. of equations it has = No. ofq qFourier coefficients C(K – G).


Chapter 7: Energy BandsSolution of the Central Equation

Eq. (27) represents a set of simultaneous linear equations thatconnect the coefficients C(k ‐ G) for all reciprocal lattice vectors G.It is a set because there are as many equations as there areIt is a set because there are as many equations as there arecoefficients C.

To solve it, the determinant of the coefficients must vanish. As an application: for the case when G = g (Shortest values of G):

2 0 0 0 ( 2 )0 0 ( )

k g gU C k gU U C k

0 0 ( )

0 0 0 (32)( )0 0 ( )

g k g g

g k g

g k g g

U U C k gU U C k

U U C k g

2

( )0 0 0 ( 2 )

g k g g

g k g

gU C k g

The solution of the determinant (32) gives a set of energy


eigenvalues nk .

Chapter 7: Energy BandsEmpty Lattice Approximation

Band structures are usually plotted as energy versus wavevectorin the first Brillouin zone. When wavevectors are outside thiszone they are carried back into the first zone by translationzone, they are carried back into the first zone by translation.

We look for a G such that a k' in the first zone satisfies: where k is the free electron wavevector in the empty lattice.

'k G k

We can drop the ‘ from k since G can be ‐ or + :2 2

2 2 2 2( , , ) ( ) ( ) ( ) ( )2 2x y z x x y y z zk k k k G k G k G k Gm m

2 2y y ym m

We consider as an example free electron bands of a simple cubiclattice. Suppose we want to exhibit the energy as a function of klattice. Suppose we want to exhibit the energy as a function of kin the [100] direction. Let . We show several bands inthis empty lattice approximation with their energies (000) at k =0 and (k 00) along the k axis in the first zone:

2 / 2 1m


0 and (kx00) along the kx axis in the first zone:

Chapter 7: Energy BandsEmpty Lattice Approximation


Nasser S. Alzayed

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